Finding points/coordinates and linear equations?

Question

In the quadrilateral ABCD, the points of A and C are (-1,4) and (2,-1) respectively. The line CD is parallel to the line with equation x+4y+8=0 and D lies on the x-axis. The foot of the perpendicular from A to BC bisects BC and rests on the x-axis with x-coordinate 3

 

a) Find coordinates of B

b) Find the equation of DC and AD

 

Super stuck on this, no understanding what so ever.

Answer 1

Hi Chanvadee, for part a), you are given the mid-point of the line BC, which is (3, 0). Using the equation to determine the mid-point of a line,

 

mid-point = ((x₁+x₂)/2, (y₁+y₂)/2), you can solve for point B.

 

Let (x₁, x₂) be represented by point C, or (2, -1), and (x₂, y₂) represent point B.

 

mid-point = (3, 0) = ((2+x₂)/2, (-1+y₂)/2)

 

Gives:

 

(2+x₂)/2 = 3, and (-1+y₂)/2 = 0

 

Solving for x₂:

 

(2+x₂)/2 = 3

 

2 + x₂ = 6

 

x₂ = 4

 

Solving for y₂:

 

(-1+y₂)/2 = 0

 

-1 + y₂ = 0

 

y₂ = 1

 

Point B is at: (4, 1)

 

Part b)

 

The equation for the line DC will have the same slope as the line x + 4y = 8 and will have an x-intercept of (4, 0).

 

The slope of line x + 4y = 8 is represented by: -A/B, where A is the coefficient of the x-term and B is the coefficient of the y-term. -A/B equals: -1/4.

 

The line, in y-intercept form is: y = (-1/4) * x + 0, or y = -1/4 * x

 

For the equation of the line AD, first find the slope using the points for A and D, (-1, 4), and (4, 0).

 

The slope of AD = (4 - 0)/(-1 - 4), or -4/5

 

Using the slope-intercept form of the line and the point, (4, 0), we can find the line for AD.

 

y - y₁ = m * (x - x₁)

 

y - 0 = -4/5 * (x - 4)

 

Multiply by 5 to clear the fraction:

 

5y = -4 * (x - 4)

 

5y = -4x + 16

 

In standard form, the line is: 4x + 5y = 16

 

Questions?